a) ko vt lại đề 

4x²-8x+x-2=0

=>4x(x-2)+(x-2)=0

=>(x-2)(4x+1)=0

......

b) bn tự làm nha

a)\(4x^2-7x-2=0\)

Ta có \(\Delta=7^2+4.4.2=81,\sqrt{\Delta}=9\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+9}{8}=2\\x=\frac{7-9}{8}=\frac{-1}{4}\end{cases}}\)

b)\(4x^2+5x-6=0\)

Ta có \(\Delta=5^2+4.4.6=121,\sqrt{\Delta}=11\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+11}{8}=\frac{3}{4}\\x=\frac{-5-11}{8}=-2\end{cases}}\)

\(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)

\(\Leftrightarrow\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)

\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{7}-\frac{4}{9}\right)=0\)

mà \(\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{7}-\frac{4}{9}\right)\ne0\)

\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)

Vậy x=-15

k cho mình nha

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left[\left(x^2-7\right)+3\right]\left[\left(x^2-7\right)-3\right]=72\)

\(\Leftrightarrow\left(x^2-7\right)^2-9=72\)

\(\Leftrightarrow\left(x^2-7\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-7=9\\x^2-7=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=16\\x^2=-2\left(loai\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

Đặt \(y=x^2-7\) Pt trở thành: \(\left(y+3\right)\left(y-3\right)=73\)

\(\Leftrightarrow y^2-9=72\)

\(\Leftrightarrow y^2=81\)

\(\Leftrightarrow y=\pm9\)

Từ đó tìm được: \(x_1=-4;x_2=4\)

\(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-\left(4x^2-8x-4\right)=0\)

\(\Leftrightarrow x^2\left(x^2+2x+1\right)-4\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}}\)

\(\text{CM vô nghiệm}\)
\(\text{a) }\left(x-2\right)^3=\left(x-2\right).\left(x^2+2x+4\right)-6\left(x-1\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6x^2+12x-6\)
\(\Leftrightarrow x^3-6x^2+12x-x^3+6x-12x=-8+8-6\)
\(\Leftrightarrow0x=-6\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{b) }4x^2-12x+10=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2=-1\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{CM vô số nghiệm}\)
       \(\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^3-3x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\text{ (luôn luôn đúng)}\)
\(\text{Vậy }S\inℝ\)

Ông Thọ, vì có một loại sữa tên là Sữa ông Thọ

Ông Thọ,sữa Ông Thọ đó.

*Chú thích:

   - Sữa Ông Thọ:Là sản phẩm sữa đặc nổi tiếng thuộc sở hữu của Công ty sữa Việt Nam(VinaMilk).Thường dùng pha café hay dùng để chấm với bánh mì(bánh mì quẹt sữa).

Chúc học tốt.Lần sau không đăng linh tinh nữa nhé ^^!

Do 13;3 là 2 số nguyên tố 

=> \(\hept{\begin{cases}m^2+mn+n^2=13k\left(1\right)\\m+2n=3k\end{cases}}\left(k\in Z\right)\)

(1) <=> \(4m^2+4mn+4n^2=52k\)

=> \(3m^2+\left(m+2n\right)^2=52k\)

=> \(3m^2+9k^2=52k\)

=> \(3m^2=k\left(52-9k\right)\)

Do \(m^2\ge0\)

=> \(0< k\le\frac{52}{9}\)

=> \(k\in\left\{1;2;3;4;5\right\}\)

+ \(k=1\)=> \(m=\sqrt{\frac{43}{3}}\left(l\right)\)

+ \(k=2\)=> \(m=\sqrt{\frac{68}{3}}\left(l\right)\)

+ \(k=3\)=> \(\orbr{\begin{cases}m=5\rightarrow n=2\\m=-5\rightarrow n=7\end{cases}}\)

+ \(k=4\)=> \(m=\sqrt{\frac{64}{3}}\left(l\right)\)

+ \(k=5\)=> \(m=\sqrt{\frac{35}{3}}\left(l\right)\)

Vậy \(\left(m,n\right)=\left(5;2\right),\left(-5;7\right)\)

\(S=a+\frac{1}{a}=\frac{8a}{9}+\left(\frac{a}{9}+\frac{1}{a}\right)\ge\frac{24}{9}+2\sqrt{\frac{a}{9}.\frac{1}{a}}=\frac{10}{3}\)

Chúc bạn học tốt !!!